Suppose you have a system with a 12 inch by 14 inch video...

Suppose you have a system with a 12 inch by 14 inch video monitor that can display $120 \mathrm{pix}-$ els per inch. If memory is organized in one byte words, the starting frame buffer address is 0 , and each pixel is assigned one byte of storage, what is the frame buffer address of the pixel with screen coordinates $(x, y)$ ?

Key Concepts

Frame Buffer

A frame buffer is a dedicated area of memory used to store the digital representation of a display image. In graphics systems, each pixel on the screen is assigned a specific location in this memory, and the complete image is built by reading the pixel values sequentially from the frame buffer and sending them to the display hardware.

Resolution and Pixel Density

Resolution refers to the number of distinct pixels in each dimension that can be displayed on the screen, while pixel density indicates how many pixels are packed into each inch of the display. By multiplying the physical dimensions of the monitor by its pixel density, you can determine the total number of pixels that make up the display’s width and height.

Pixel Coordinates and Memory Addressing

Pixel coordinates refer to the (x, y) positions on the display. Memory addressing in a frame buffer is typically organized such that each pixel corresponds to a byte (or group of bytes) in memory. The location corresponding to a given (x, y) coordinate is calculated by considering the width of the display and the sequential storage of rows, often using a formula that multiplies the y-coordinate by the number of pixels per row before adding the x-coordinate.

Row-Major Order Memory Organization

Row-major order is a method for storing multidimensional arrays in which the complete first row is stored in memory, followed by the complete second row, and so on. This concept is fundamental in frame buffer addressing, as it allows the calculation of a pixel's memory address by first going to the correct row (using a multiplication by the row width) and then moving to the appropriate column (addition of the x offset), thereby providing efficient access to each pixel’s data.

Transcript

00:01 Okay, here we go with this question.

00:03 We've got four megabytes of main memory, and each cell is going to be holding one byte of information.

00:11 So i drew on the right, i drew a little sample of what main memory will look like.

00:15 Our first cell is always going to be zero, zero.

00:18 That's where we start counting, and we stop counting at the hexadecimal representation for 15.

00:24 That's f.

00:25 So in this case, using two digits, we can go from zero to 255, which leaves us with 256 bytes used or 256 cells.

00:36 Now that's pretty good, 256 bytes, but this one's going to be a lot bigger.

00:40 This is 4 million bytes.

00:43 So let's get started with this.

00:46 We need to take that number 4 million and divide it down into hexadecimal representation.

00:52 Actually, what i'm going to do is i'm going to start with this number.

00:56 It is one less than 1 million, one less than 4 million.

00:59 The reason i'm doing this is because we need to remember our first cell is 0 -0.

01:07 So that's important to start with.

01:10 Divide by 16 to get our hexadecimal representation.

01:14 If you've got your calculator, you can punch this in and see your answer is 249 ,999.

01:23 And here is our remainder.

01:26 In a fraction form is 1516s.

01:28 Now what is hexadecimal representation of 15? it's f right here...

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